Ok, after some Google searches, reading and remembering my RF studies I think I am following your equations.
In: 10log(KT)+10log(RBW=IF filter of S.A)+ NF of S.A
The first term is “minimum equivalent input noise for the receiver” – 10log(k*T) = 10log( (1.38 x 10^-23 J/K * 290°K)*1000) = -174dBm/Hz
This assumes an ambient temperature of 290°K. If the temperature is different then the result is different.
the second term is ‘noise passed’ and is Bandwidth limited – 10log(BW) = 10log(1000Hz) = 30dB
Third is the NF of analyzer – NF = 30dB (this should be in the NA’s specs)
So, putting together the numbers: Noise level = -174dBm + 30dB + 30dB = -114dB. This looks very close to what you show in the first picture in your link, -112dBm is only 2dB off the calculation…
So now you add an Amplifier. So: -114dB + 3dB + 30dB = -81dB which is not what you see in the second picture so this equation does not work or some number are wrong.
What you need to do is calculate the Noise Figure for Cascaded Stages.
http://www.ece.ucsb.edu/yuegroup/Teachi … basics.pdf
http://www.dsplog.com/2012/04/21/noise- … ed-stages/
The Noise Factor of the SA = 1000, the LNA = 2
Gain of LNA = 30dB = 1000
Noise Factor = 2 + [(1000-1)/1000] = 2.999 (round to 3.0)
the Noise Figure is now 10log(3) = 4.77dB
What this means is that the SA has almost NO contribution to the overall noise figure and also shows the importance of a low NF in the first RF stage.
Now go back and combine these number into the first equation:
Noise level = -174dBm + 30 + 30 + 4.77 = -109dBm which is only a few dB different than your second picture.
Does this now make sense?